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Integration by Substitution

  1. Nov 30, 2004 #1
    I'm stuck on how to advance further on this problem and if anyone can point my in the right direction I would be greatly appreciative.

    [tex]\int\frac{dx}{\sqrt{x(1-x)}}[/tex]

    The integral has to be solved using substitution, but we are required to use
    [tex]u=\sqrt{x}[/tex]

    From this:
    [tex]du=\frac{dx}{2\sqrt{x}}[/tex]

    But I am stuck on how to convert the remaining portion of the function in terms of du.
    [tex]\int\frac{dx}{u\sqrt{1-x}}[/tex]
     
  2. jcsd
  3. Nov 30, 2004 #2

    ShawnD

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    I gave it a try and couldn't get anywhere with it. Maple says the answer is arcsin(2x-1).

    Is that exactly how the question was given?
     
  4. Nov 30, 2004 #3
    [tex]u = \sqrt{x}[/tex]

    so

    [tex]u^2 = x[/tex]

    and

    [tex]2du = \frac{dx}{\sqrt{x}}[/tex]

    First use the third equation, then use the second equation to get rid of any other instances of x that're left.

    And Shawn is not correct in his solution.

    --J
     
    Last edited: Nov 30, 2004
  5. Dec 1, 2004 #4
    Shaun's solution looks good to me, what do you propose the actual answer is Justin?
     
  6. Dec 1, 2004 #5
    Complete the square within the square root in the denominator and the apply the result

    [tex]\int\frac{dx}{\sqrt{a^2-x^2}} = arcsin\frac{x}{a} [/tex]

    spacetime
    www.geocities.com/physics_all
     
  7. Dec 1, 2004 #6
    [tex]\int\frac{dx}{\sqrt{x(1-x)}} = 2 \arcsin{\left(\sqrt{x}\right)}[/tex]

    Differentiate it and you'll get the integrand.

    The derivative of arcsin(2x-1) is [tex]\frac{2}{\sqrt{4x^2 - 4x + 2}}[/tex].

    --J
     
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